Tanya Khovanova's Math Blog

This year I am again on the organizing committee of the Women and Mathematics program at Princeton’s Institute for Advanced Study. Our subject is “p-adic Langlands Program.” It is a fashionable, advanced and very influential program connecting number theory and representation theory.

We invite undergraduate students, graduate students and post-docs to apply. In 2009-2010 the Institute has been running a special year in Analytic Number Theory. That has brought many number theorists to the institute already, so there will be a lot of people to talk to.

Last year I promised to hold a math party during the program. But I had to cancel it due to a scheduling conflict with George Hart’s ZomeTool Workshop. I am planning a party this year. Either way, we’ll have fun.

If you want to learn about the Langlands program, to spent time on the beautiful grounds of the Institute, to eat in one of the best cafeterias around, and to make new friends with other women interested in number theory, then please apply. The application deadline is February 20.

Suppose you want to increase your chances of winning the lottery jackpot by pooling money with a group of coworkers. There are several issues you should keep in mind.

When you pool the money and you hit the jackpot, the money has to be split. If you bought 10,000 tickets and the jackpot that you win is $100 million, then each ticket is entitled to a mere $10,000. Your chances of hitting the jackpot in the first place are 1 in 17,500 and you’re not going to get rich off what you win.

Perhaps you’d be satisfied with a small profit. However, as I calculated in my previous piece on the subject, even if you include the jackpot in the calculation of the expected return, the Mega Millions game never had, and probably never will have a positive return.

Despite this fact, people continue to pool money in the hopes of winning big. However, there are more problems in doing this than just its non-profitability.

Consider a scenario. Your coworkers collected $1,000 to buy 1,000 lottery tickets. You give the money to Jerry who buys the tickets. Jerry can go to a store and buy 1,005 tickets. After the lottery he checks the tickets, takes the best five for himself and comes back to work with 1,000 disappointing tickets.

It is more likely that Jerry is cheating or that he will lose the tickets than it is that your group will win the jackpot. But there is a probabilistic way to check Jerry’s integrity. According to the odds, every 40th ticket in Mega Millions wins something. Out of 1,000 tickets that Jerry bought, you should have about 25 that win something. If Jerry systematically brings back tickets that win less often than expected, you should replace Jerry with someone else.

There are methods to protect your group against cheating. For example, you can ask another person to join Jerry in purchasing the tickets, which they then seal in an envelope that they both sign.

Alternatively, you yourself could be the person responsible for buying 1,000 tickets. How would you protect yourself from suspicion of cheating? The same way as I mentioned above: bring along some witnesses and have everyone sign the sealed envelope.

The most reliable way to prevent Jerry from cheating is to have him write down all the ticket numbers and send this information to everyone before the drawing. This way he can’t replace one ticket with another. But this is a lot of work for tickets that are usually worth less than the money you collected to buy them.

But there are other kinds of dangers if you use this supposedly reliable method. If you bought a lot of tickets the probability of winning a big payoff increases. Suppose Jerry publicly locks the envelope in a desk drawer in his office. If one ticket wins $10,000, and everyone knows all the ticket combinations, suddenly Jerry’s desk drawer becomes a very unsafe place to keep the tickets.

Scams are not your only worry. You shouldn’t buy the same combination twice — whether picking randomly or not. You really do not want to waste a ticket and end up sharing the jackpot with yourself.

You cannot change the odds of hitting the jackpot, but you can change the odds of sharing it with others. Indeed, there are people who do not buy random combinations, but rather pick their favorite numbers, like birthdays. You can reduce the probability of sharing the jackpot if you choose the combinations for your tickets wisely, by picking numbers that other people are unlikely to pick.

Still want to try the lottery? If you feel a need to throw your money away, instead of buying lottery tickets, feel free to donate to this blog.

In one of my previous pieces, I discussed returns on the Mega Millions lottery game, assuming that you buy a small number of tickets. In such a case winning the jackpot has zero probability. So I argued that if you want to estimate the profitability of the lottery as an investment, you have to remove the jackpot money from the calculation.

Today I will discuss what the formal expected return is. That is, I will include the jackpot money in the calculation. Since I argued against including the jackpot in my last article, you might wonder why I’ve then turned around to look into this.

I think this mathematical exercise will be fun. Besides, on a practical note, it is useful to know when the formal expected return is more than 100%, because then it might make sense to pool money with other people. Keep in mind though that if you want a chance to hit the jackpot, the total number of tickets you buy must be really big. For example, even if you manage to pool $10,000 for tickets, your probability of winning the jackpot in Mega Millions is only one in 17,500 — still minuscule.

If you buy only one ticket, you’ll lose. If you manage to pool a lot of money and the probability of the jackpot becomes noticeable, that is, non-zero, could the jackpot be large enough that the lottery becomes a good investment?

For this calculation, I’m still assuming that you buy a relatively small number of tickets. If you buy millions of tickets the calculation is slightly different, and I will write about that later.

You might think that when the jackpot is bigger than the odds, it makes sense to play. I am discussing the Mega Millions game, where the odds of winning the jackpot are one in 175 million. So if the jackpot is more than $175 million, then it is profitable to play. Right?

Wrong. As I mentioned in my previous piece, after reducing for taxes, you get about 16% of your money back through smaller payouts. Hence, you need to recover the other 84% through the jackpot. So the jackpot should be more than 175*.84 = 147 million dollars. This sounds even better. Right?

Wrong. No one receives the jackpot. Winners can chose to immediately receive the lump sum, which equals the money lottery organizers have actually set aside for it. Alternatively, the lottery organizers can invest the lump sum and give winners a yearly distribution over many years, the total of which will equal the jackpot.

Suppose for simplicity the lump sum is half of the jackpot. That means we need the jackpot to be $294 million ($147 x 2). Right?

Oops. As usual, we forgot about taxes. To exacerbate your pain, I have to add that the winnings are taxable. Suppose you have to pay 30% from the jackpot. That means the jackpot needs to be $424 ($294/0.7) million in order to justify pooling money. OK?

We haven’t seen jackpots that big yet. But neither have we finished the calculation. There is a probability that you might have to share the jackpot with other winners. To calculate this probability, we need to calculate the number of tickets sold. That means, your expected return depends not only on the size of the jackpot, but also on the number of these tickets.

But even if you know the number of tickets sold, we cannot calculate the expected returns precisely because people don’t always buy tickets with random combinations, but often pick their own numbers.

When the jackpot is large people start buying tons of tickets, so we can expect that many of them buy quick-picks. Let us assume for now that the vast majority of people do not choose their own numbers, but buy tickets at random. Suppose 200 million tickets were sold. That is a very big number. Last time that many tickets were sold was when the jackpot was $390 million in March 2007. By the way, that was the largest jackpot ever.

In order to finish the calculation, we need to establish the probability of several winners, given that 200 million random tickets were sold:

From here we can calculate the adjustment coefficient, that is, the proportion of money you are expected to get from the jackpot given that there are 200 million players in the game. The coefficient is calculated from the table above as (0.3647 + 1/2*0.2075 + 1/3*0.0787 +1/4*0.0224 + 1/5*0.0051 + 1/6*0.0010)/(1 – 0.3204), and is equal to 0.7379. We need to divide our previous figure of $424 million by the adjustment coefficient. The result is $575 million.

Given that a $390 million jackpot attracted more than $200 million in tickets, we can expect that the $575 million jackpot will make people completely crazy and attract even more money. So I do not anticipate that the Mega Millions game will ever have a positive formal expected gain. My conclusion is that not only is there no financial sense in buying a single lottery ticket, but also none in pooling money.

Of course, you can buy tickets for non-financial reasons, like pumping up your adrenaline. In any case, I showed you the method to calculate your expected return, or, more appropriately, your expected loss.

Here is a fun math activity I use with my students, after I teach them to play the game of set. To other teachers — feel free to clone this idea.

First, I ask the students if they know what a magic square is. They usually do know that a magic square is a three-by-three square of distinct digits, so that every row, column and diagonal has the same sum. Then I ask them what a magic set square might be. Often they guess correctly that it is a three-by-three square made of set cards, so that every row, column and diagonal form a set. Once that’s established, I have them build magic set squares.

While they’re building them, I ask a lot of questions, from how many cards there should be in the deck to how many different sets there are.

Once the squares are built, I ask them what a magic set cube might be. Their next task is to use their magic set squares as the bottom layer in building magic set cubes. In order to see all the cards in the cube, I instruct them to arrange the layers (bottom, middle and top) side-by-side.

As they’re working on their cubes, I continue quizzing them. How many main diagonals does a cube have? Once they confirm that the answer is four, I ask them to show me those diagonals in their magic set cubes and check that they are sets. I might also ask them how many different magic set squares should be inside a magic cube. This is a theoretical math question they need to answer before finding them in their own model. Next they need to identify the different sets that form lines inside their cubes.

At this point, some students guess my next request: to construct a magic set hypercube.

After students build their hypercubes, they never want to destroy them. They like comparing the different hypercubes and often take photos of them. If there’s still time left, I can continue in several directions. For example, they can count the main diagonals of the hypercube and find them in their models. Alternatively, they can find a “no set” — the largest possible set of cards inside a magic set hypercube that doesn’t contain a set.

Math is usually about thinking, but this is one activity the students can do with their hands. And that adds another layer of magic.

I promised to discuss how to improve the accuracy of your guessing at AMC 10/12, or other tests for that matter. There are two types of guessing. First, meta-guessing is when you do not look at the problem, but rather guess from just looking at the choices. Second, in the one I refer to as an educated-guess you do look at the problem. Instead of completely solving it, you try to deduce some information from the problem that will help you eliminate some of the choices. In this essay, I discuss both methods.

But first let me emphasize that solving problems is a more important skill than meta-guessing from a list of choices. Solving problems not only teaches you to think mathematically, but also increases your brain power. Spending time improving your meta-guessing skills can help you at multiple-choice tests and may give you insight into how test designers think, but this will not increase your math knowledge in the long run.

On the other hand, educated-guessing is a very useful skill to obtain. Not only can it improve your score during a test, the same methods can be applied to speed up the process of re-checking your answers before handing in the test. This skill will also come in handy when you start your research. Some problems in research are so difficult that even minor progress in estimating or describing your answer is beneficial.

Before discussing particular methods, let me remind you that AMC 10/12 is a multiple-choice competition with five choices for each question. The correct answer brings you 6 points. A wrong answer brings you 0 points, and not answering brings you 1.5 points. So if you randomly guess one of the five choices, your expected average score is 1.2 points, which is 0.3 less than your score for an unanswered question. Thus guessing is unprofitable on average.

However, if you can eliminate one choice, your expected average score becomes 1.5 points. In this case guessing doesn’t bring you points on average, but it does create some randomness in your results. For strategic reasons, you might prefer guessing, as I discussed in my earlier piece.

If you can eliminate two wrong choices, then guessing becomes profitable. A random guess out of three possibilities brings you 2 points, a better result than 1.5 points for an unanswered question. Even more, if you can eliminate three choices, then guessing will increase your score by 3 points on average.

Now that we’ve covered the benefits of excluding choices before guessing, I would like to discuss how to exclude choices by just looking at them. Let us take one of the problems from the 2002 AMC10B. Here are the choices: (-2,1), (-1,2), (1,-2), (2,-1), (4,4). The pair (4,4) is a clear outlier. I suggest that an outlier can’t be a correct choice. If (4,4) were the correct answer, then it would have been enough, instead of solving the problem, to use some intermediate arguments to choose it. For example, if you can argue that both numbers in the answer must be at least 2, or must be positive or be even, then you can get the correct answer without solving the problem. Any problem for which you can easily pick the correct answer without solving it is an unacceptably poor problem design. Thus, (4,4) can’t be the correct answer, and should be eliminated during guessing.

Let us look at a 2002 AMC10A problem with the following choices: 4/9, 2/3, 5/6, 3/2, 9/4. Test designers want to create choices that are plausible. They try to anticipate possible mistakes. In this set of choices, we can deduce that one of the mistakes that they anticipate is that students will confuse a number with its inverse. In this case 5/6 can’t be the correct answer. Otherwise, 6/5 would have been included as a choice. In another similar example from the 2000 AMC10 with choices -2, -1/2, 1/3, 1/2, 2, the designers probably hope that students will confuse numbers with their inverses and negations. Hence, we can exclude 1/3.

Sometimes the outlier might hint at the correct answer. Suppose you have the following list of choices: 2, 1/2π, π, 2π, 4π. The number 2 is an outlier here. Probably, the problem designers were contemplating that students might forget to multiply by π. In this case the likely correct answer would be 2π, because only from this answer can you get 2 by forgetting to multiply by π.

As an exercise, try to eliminate the wrong choices from the following set from a problem given at 2000 AMC10: 1/(2m+1), m, 1-m, 1/4m, 1/8m².

AMC designers know all of these guessing tricks, so they attempt to confuse the competitors from time to time by going against common sense. For example, in a 2002 AMC10A problem the choices were: -5, -10/3, -7/3, 5/3, 5. I would argue that -7/3 is a clear outlier because all the others are divisible by 5. Furthermore, there is no point in including so many answers with 3 in the denominator unless there is a 3 in the denominator of the correct answer. So I would suggest that one of -10/3 and 5/3 must be the answer. My choice would be 5/3, as there is a choice of 5 which I would assume is there for students who forgot to divide by 3. As I said the designers are smart and the actual answer is -10/3. They would have tricked me on this problem.

One of the best ways to design a multiple choice question is to have an arithmetic progression as a list of choices. There is no good way to eliminate some choices from 112, 113, 114, 115, 116. Unfortunately for people who want to get an advantage by guessing, many of the problems at AMC have an arithmetic progression as their choices.

Now that we’ve seen the methods of meta-guessing, let’s look at how to make an educated guess. Let us look at problem 1 in 2003 AMC10A.

What is the difference between the sum of the first 2003 even counting numbers and the sum of the first 2003 odd counting numbers?

Without calculations we know that the answer must be odd. Thus, we can immediately exclude three out of five choices from the given choices of 0, 1, 2, 2003, 4006. Parity consideration is a powerful tool in eliminating wrong answers. Almost always you can decide the parity of the answer much faster than you can calculate the answer. Similar to using parity, you can use divisibility by other numbers to have a fast elimination. Here is a problem from 2000 AMC10:

Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71, 76, 80, 82, and 91. What was the last score Mrs. Walter entered?

The first four numbers entered must be divisible by 4. The total of the given numbers is divisible by 4. Hence, the last number must also be divisible by 4. This reasoning eliminates three out of five choices.

Another powerful method is a rough estimate. Let us look at the next problem in 2003 AMC10A.

Members of the Rockham Soccer League buy socks and T-shirts. Socks cost $4 per pair and each T-shirt costs $5 more than a pair of socks. Each member needs one pair of socks and a shirt for home games and another pair of socks and a shirt for away games. If the total cost is $2366, how many members are in the League?

If we notice that the cost per person is more than $25, we can conclude that there are less than a hundred members in the League. Given the choices of 77, 91, 143, 182, and 286, we immediately can eliminate three of them.

Another method is to use any partial knowledge that you may have. Consider this problem from 2003 AMC10A:

Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected?

You might remember that there is a formula for this. Even if you do not remember the exact formula, you might still have a vague memory that the answer must be a binomial coefficient that somehow uses the number of cookies and the number of flavors. Looking at the choices — 22, 25, 27, 28, 29 — you can see that the only choice that appears in the first 10 rows of the Pascal’s triangle is 28. So you should go with 28.

It is easy to talk about easy problems; let us see what we can do about difficult ones. Consider the last problem on 2003 AMC10A:

Let n be a 5-digit number, and let q and r be the quotient and the remainder, respectively, when n is divided by 100. For how many values of n is q+r divisible by 11?

The choices are 8180, 8181, 8182, 9000, 9090. They can be naturally split into two groups: three choices below 9000 and the rest. By my rules of removing outliers the group of numbers below 9000 seems the more promising group. But I would like to discuss how to approximate the answer. There is no reason to believe that there is much correlation between remainders by 100 and divisibility by 11. There is a total of 90,000 5-digit numbers; among those numbers, approximately 90,000/11 = 8182 is divisible by 11, so we should go with the group of answers close to 8182.

Another way of thinking about this problem is the following. There are 900 different quotients by 100 to which we add numbers between 0 and 99. Thus for every quotient our sums are a set of 100 consecutive numbers. Out of 100 consecutive numbers usually 9, and rarely 10, are divisible by 11. Hence, the answer has to be less than 9000.

Sometimes methods you use for guessing can bring you the answer. Here is a problem from 2001 AMC12:

What is the product of all odd positive integers less than 10000? (A) 10000!/5000!², (B) 10000!/2⁵⁰⁰⁰, (C) 9999!/2⁵⁰⁰⁰, (D) 10000!/(2⁵⁰⁰⁰5000!), (E) 5000!/2⁵⁰⁰⁰.

For a rough estimate, I would take a prime number and see in what power it belongs to the answer. It’s simplest to consider a prime number p that is slightly below 5000. Then p should appear as a factor in the product of all odd positive integers below 10000 exactly once. Now let us look at the choices. Number p appears in 5000! once and in 10000! twice (as p and 2p). Hence, it appears in (A) zero times, and twice each in (B) and (C). We also can rule out (E) as the product of odd numbers below 10000 must be divisible by primes between 5000 and 10000, but 5000! doesn’t contain such primes. Thus the answer must be (D).

The method I just described won’t produce the formula. But the ideas in this method allow you to eliminate all the choices except the right one. Moreover, this method provides you with a sanity check after you derive the formula. It also helps to build your mathematical intuition.

I hope that you will find my essays about AMC useful. And good luck on February 9!

I teach students to solve math problems by appreciating the big picture, or by noticing the problem’s inner symmetries, or through a deep understanding of the problem. In the long run, one thing leads to another: such training structures their minds so that they are better at understanding mathematics and, as a consequence, they perform well at math competitions.

That is why, when AMC is still far away, I do not give my students a lot of AMC problems; rather, I pick problems that contain useful ideas. When I do give AMC problems, I remove the multiple choices, so they understand the problems completely, instead of looking for shortcuts. For example, this problem from AHSME 1999 is a useful problem with or without choices.

What is the largest number of acute angles that a hexagon can have?

As AMC approaches, we start discussing how to solve problems given multiple choices. Training students for AMC is noticeably different from teaching mathematics. For example, some problems are very specific to AMC. They might not even exist without choices. Consider this problem from the 2001 AMC12:

A polynomial of degree four with leading coefficient 1 and integer coefficients has two zeros, both of which are integers. Which of the following can also be a zero of the polynomial?

Here we really need the choices in order to pick one complex number, such that its real part is an integer or a half integer and, in addition, the product of the number with its conjugate produces an integer.

Sometimes the choices distract from solving the problem. For example, in the following problem from the 2005 AMC12, having choices might tempt students to try to eliminate them one by one:

The sum of four two-digit numbers is 221. None of the eight digits is 0 and no two of them are same. Which of the following is not included among the eight digits?

Without the choices, students might start considering divisibility by 9 right away.

On some occasions, the choices given for the problems at AMC make the problem more interesting. Here is an example from the 2000 AMC10:

Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?

The choices are 21, 60, 119, 180 and 231. We can immediately see that the answer must be odd. Because the span of the three remaining choices is so wide, we suspect that we can eliminate the smallest and the largest. Trying for 5 and 7 — the two smallest primes in the range — we can eliminate 21. Similarly, checking the two largest primes in the range, we can eliminate 231. This leaves us with the answer: 119. If the choices were different, we might have lost the interplay between the solution and the list of choices. Then, solving the problem would have been slower and more boring. There are ten pairs of prime numbers to check. And we would need on average to check five of them until we stumbled on the correct choice.

In other cases having multiple choices makes the problem more boring and less educational. Here is another problem from the same competition.

Two non-zero real numbers, a and b, satisfy ab = a – b. Find a possible value of a/b + b/a – ab.

Solving this problem without choices can teach students some clever tricks that people use when playing with expressions. Indeed, when we collect a/b + b/a into one fraction (a² + b²)/ab, we might remember that a² + b² is very close to (a – b)², and see from here that a/b + b/a – 2 is (a – b)²/ab, which, given the initial condition, equals ab. Thus, we can get the answer: 2.

On the other hand, if you look at the multiple choices first: -2, -1/2, 1/3, 1/2, 2, you might correctly assume that the answer is a number. Thus, the fastest way to solve it is to find an example. If a = 1, then b must be 1/2, and the answer must be 2. This solution doesn’t teach us anything new or interesting.

My next example from the 2002 AMC10 is similar to the previous one. The difference is that the solution with multiple choices is even more boring, while the solution without these choices is more interesting and beautiful.

Let a, b, and c be real numbers such that a – 7b + 8c = 4 and 8a + 4b – c = 7. Then a² – b² + c² is: 0, 1, 4, 7, or 8.

Given the choices, we see that the answer is a number. Hence we need to find any solution for the system or equations: a – 7b + 8c = 4 and 8a + 4b – c = 7. For example, if we let c = 0, we have two linear equations and two variables a and b that can be solved by a straightforward computation. Then we plug the solution into a² – b² + c².

Without knowing that the result is a number, we need to look at the symmetries of our two initial equations. We might discover a new rule:

If we have two expressions ax + by and bx – ay, where a and b are switched between variables and there is a change in sign, it is a good idea to square each of them and sum them up, because the result is very simple: (a² + b²)(x² + y²).

Hence, in our initial problem we need to move the term with b in our two linear equations to the right; then square them and sum up the results. This way we may get a very simple expression. And indeed, this trick leads to a solution and this solution provides insight into working with algebraic expressions.

This is the perfect problem to linger over, assuming you’re not in the middle of a timed competition. It might make you wonder for which parameters this problem works. You might discover a new theorem that allows you to create a very similar problem from any number that can be represented as a sum of two non-trivial squares in two different ways.

To prepare my students for AMC I need to teach them tricks that are not useful at USAMO, or in mathematics in general for that matter. Many tricks distract from new ideas or from understanding the problem. All they give us is speed.

This bothers me, but to pacify myself, I keep in mind that most of my students will not become mathematicians and it might be useful in their lives to be able to make split-second decisions among a small number of choices.

However, it seems like Americans have the opposite problem: we make quick decisions without thinking. I’m concerned that training for multiple choice tests and AMC competitions aggravates this problem.

By Tanya Khovanova and Gregory Bomash

What object is missing?

Should you try to guess an answer to a multiple-choice problem during a test? How many problems should you try to guess? I will talk about the art of boosting your guessing accuracy in a later essay. Now I would like to discuss whether it makes sense to pick a random answer for a problem at AMC 10.

Let me remind you that each of the 25 problems on the AMC 10 test provides five choices. A correct answer brings you 6 points, a wrong answer 0 points and not answering at all gives you 1.5 points. So guessing makes the expected average per problem to be 1.2 points. That is, on average you lose 0.3 points when guessing. However, if you are lucky, guessing will gain you 4.5 points per problem, and if you are unlucky, it will lose you 1.5 points per problem.

So we see that on average guessing is unprofitable. But there are situations in which you have nothing to lose if you get a smaller score and a lot to gain if you get a better score. Usually the goal of a competitor at AMC 10 is to get to AIME. For that to happen, you need to get 120 points or be in the highest one percentile of all competitors. This rule complicates my calculations. So I decided to simplify it and say that your goal is to get 120 points and then see what mathematical results I can get out of that simplification.

First, suppose you are so accurate that you never make mistakes. If you have solved 20 problems, then your score without guessing is 127.5. If you start guessing and guess wrongly for all of the last five questions, you still have your desired 120 points. In this instance it doesn’t matter whether you guess or not.

Suppose on the other hand that you are still accurate, but less powerful. You have only solved 15 problems, so your score without guessing is 105. Now you must be strategic. Your only chance to get to your goal of 120 points is to guess. Suppose you randomly guess the answers for the 10 problems you didn’t solve. To make it to 120, you need to guess correctly at least five out of the ten remaining problems. The probability of doing so is 3%. Here is a table of your probability of making 120 points if you solve correctly n problems and guess the other problems.

We can see that if you solved a small number of problems, then the probability of getting 120 points is minuscule; but as the number of problems you solved increases, so does the probability of getting 120 points by guessing.

The interesting part is that if you have solved 19 problems, you are guaranteed to get to AIME without guessing. On the other hand, if you start guessing and all your guesses are wrong, you will not pass the 120 mark. The probability of having all six problems wrong is a not insignificant 26%. In conclusion, if you are an accurate solver and want to have 120 points, it is beneficial to guess the remaining problems if you solved fewer than 19 problems. It doesn’t matter much if you solved fewer than 10 problems or more than 19. But you shouldn’t guess if you solved exactly 19.

If you are not 100% accurate things get more complicated and more interesting. To decide about guessing, it is crucial to have a good estimate of how many mistakes you usually make. Let’s say that you usually have two problems wrong per AMC test. Suppose you gave answers to 20 problems at AMC. What’s next? Let us estimate your score. Out of your 20 answers you are expected to get 18*6 points for them plus 5*1.5 points for the problems you didn’t answer. Your expected score is 115.5. You are almost there. You definitely should guess. But does it matter how many questions you are trying to guess?

The correct answer to one question increases your score by 4.5 with probability 0.2, and the wrong answer decreases your score by 1.5 with probability 0.8. One increase by 4.5 is enough for you goal of 120 points. So if you guess one question, with probability 0.2, you get 120 points. If you guess two questions, then your outcome is as follows: you increase your score by 9 points with probability 0.04; you increase your score by 3 points with probability 0.32; and you decrease your score by 3 points with probability 0.64. As you need at least a 4.5 increase in points, it is not enough to guess one question out of two. You actually need to guess correctly on both of them. The probability of this happening is 0.04. It is interesting, but you have a much greater chance to get to your goal if you guess just one question than if you guess two. Overall, here is the table of probabilities to get to 120 points where m is the number of questions you are guessing.

Your best chances are to guess all the remaining questions.

By the end of the test you know how many questions you answered, but you don’t know how many errors you made. The table below tells you what you need in order to get 120 points. Here is how you read the table: The number of problems you solved is in the first column. If you are sure that the number of mistakes is not more than the number in the second column, you can relax as you made at least 120 points. The last column gives the score.

If the number of mistakes you made is one more than in the corresponding row of the table, you should start guessing in order to try to get 120 points. Keep in mind that there is a risk: if you are not sure how many problems you solved already and start guessing, you might ruin your achievement of 120 points.

In the next table I show how many questions you can guess without the risk of going below 120 points. The word “all” means that it is safe to guess all the remaining questions.

You can see that if your goal is to get 120 points, your dividing line is answering 22 questions. If you solved 22 questions or more, there is no risk in guessing. Namely, if you have already achieved more than 120 points, guessing will not take you below that. But if you made more errors than are in the table, then guessing might be beneficial. Hence, you should always guess in this case — you have nothing to lose.

Now I would like to show you my calculations for a situation in which you are close to 120 points and need to determine the optimum number of questions to guess. The first column is the number of answered questions. The second column is the number of mistakes. The third column is your expected score without guessing. The fourth column is the optimum number of questions you should guess. And the last column lists your chances to get 120 points if you guess the number of questions in the fourth column.

You can see that almost always if you are behind your goal, you should try to guess all of the remaining questions, with one exception: if you answered 19 questions and one of them is wrong. In this case you should guess exactly two questions — not all that remain.

Keep in mind that all these calculations are very interesting, but don’t necessarily apply directly to AMC 10, because I simplified assumptions about your goals. It may not be directly applicable, but I hope I have expanded your perspective about how you can use math to help you understand how better to succeed at math tests and how to design your strategy.

I plan to teach you how to guess more profitably, and this skill will also advance your perspective.

Lottery is a tax on people bad at math.

In this article I calculate how bad the lottery is as an investment, using Mega Millions as an example. To play the game, a player pays $1.00 and picks five numbers from 1 to 56 (white balls) and one additional number from 1 to 46 (the Mega Ball number, a yellow ball).

During the drawing, five white balls out of 56 are picked randomly, and, likewise, one yellow ball out of 46 is also picked independently at random. The winnings depend on how many numbers out of the ones that a player picks coincide with the numbers on the balls that have been drawn.

So what is your expected gain if you buy a ticket? We know that only half of the money goes to payouts. Can you conclude that your return is 50%?

The answer is no. The mathematical expectation of every game is different. It depends on the jackpot and the number of players. The more players, the bigger is the probability that the jackpot will be split.

Every Mega Millions playslip has odds printed on the back side. The odds of hitting the jackpot are 1 in 175,711,536. This number is easy to calculate: it is (56 choose 5) times 46.

How much is 175,711,536? Let’s try a comparison. The government estimates that in the US we have 1.3 deaths per 100 million vehicle miles. If you drive one mile to buy a ticket and one mile back, your probability to die is 2.6/100,000,000. The probability of dying in a car accident while you drive one mile to buy a lottery ticket is five times higher than the probability of winning the jackpot.

Suppose you buy 100 tickets twice a week. That is, you spend $10,000 a year. You will need to live for 1,000 years in order to make your chances of winning the jackpot be one out of 10. For all practical purposes, the chance of winning the jackpot are zero.

As the probability of winning the jackpot is zero, we do not need to include it in our estimate of the expected return. If you count all other payouts then you are likely to get back 18 cents for every dollar you invest. You are guaranteed to lose 82% of your money. If you spend $1000 a year on lottery tickets, on average you will lose $820 every year.

If you do not buy a lot of tickets your probability of a big win is close to zero. For example, the probability of winning $250,000 (that is guessing all white balls, and not guessing a yellow ball) by buying one ticket is about 1 in 4 million. The probability of winning $10,000 — the next largest win — is close to 1 in 700,000. If we say that you have no chance at these winnings anyway, then your expected return is even less: it is 10 cents per every dollar you invest.

You might ask what happens if we pool our money together. When a lot of tickets are bought then the probability of winning the jackpot stops being zero. I will write about this topic later. For now this is what I would like you to remember. From every dollar ticket:

• 50 cents goes to the state
• 32 cents towards the jackpot
• 18 cents to other winners

I am not at all trying to persuade you not to buy tickets. Lottery tickets have some entertainment value: they allow you to briefly dream about what you would do with those millions of dollars. But I am trying to persuade you not to buy lottery as an investment and not to put more hope into it than it deserves. If you treat lottery tickets as tickets to a movie that is played in your head, you will never buy more than one ticket at a time.

That is it. I advise you not to buy more than one ticket at a time. One ticket will allow you to dream about the expression on your sister’s face when she sees your new $5,000,000 mansion, but will not destroy your finances.

Should you allocate time for checking your answers during important tests? I will use AMC 10/12 as an example, but you can adjust the calculations for any other test.

AMC 10/12 is a math competition that asks 25 multiple-choice questions that you need to solve in 75 minutes. You get 6 points for a correct answer, 1.5 points for an unanswered question and 0 points for a wrong answer.

Whether or not to take time to check your answers depends on you and the situation. If you finished your test and have some time left over, then surely you should use the extra time for checking your answers. If you only have three minutes left and your next problems are too complex to be dealt with in that time, then it is logical to use these moments to check back.

Sometimes, though, it isn’t worth it to check your answers. If you haven’t finished the test, but are a super-accurate person and never make mistakes, then it is better to continue working on the next problem than to waste time checking your correct answers. Also, if you rarely catch your own mistakes anyway, it doesn’t make sense to check.

But things are not usually so clear. By the end of the test, most people need to make a decision: continue working through the problems or use the final moments to check the answers? How can you best decide if you should allocate time for checking and, if so, how much time?

The problems in AMC tests increase in difficulty. I suggest that each time you take the test or practice for AMC, take note of two things. How long did it take you to solve the test’s last problem and what is the level of your accuracy for it. Suppose you know that at the end of the AMC test you can solve a problem in about 10 minutes and it is correct about 90% of the time. That means that investing the last ten minutes in solving the next problem will give you on average 5.4 points. If you remember that a blank answer gives you 1.5 points, you should realize that solving the last problem increases your score by 3.9 points. If you are very accurate, your score can increase more, but not more than 4.5 points.

Try conducting the following experiment. Take an AMC test from a past year. Do it for 65 minutes — the time of the test minus the time you need for that last problem. Then spend the last 10 minutes checking and correcting your answers. Now let us calculate how profitable that would be. Compare the scores you would have gotten without your corrections and with your corrections. If checking increases your score by more than 3.9 points, it is more profitable to check than to solve the next problem. If you do not make errors when you’re trying to make corrections, the rule of thumb is that correcting one mistake is better than solving one problem. Indeed, your score increases by 6 points if you correct a mistake, and by not more than 4.5 points if you solve the next problem.

On all tests that punish wrong answers, correcting an error produces more points than solving a new question.

If you find that checking is profitable, but you can’t check all the problems in ten minutes, you should consider allocating more time. Keep in mind, though, that you should adjust the sample calculation above for the last two problems. Remember, the next to the last problem is generally easier than the last problem. So if it takes you ten minutes to solve the last problem in the test, it most probably will take you less than twenty minutes to solve the last two. Also, since the difficulty increases throughout the test, the accuracy of the second to last problem might be better than the accuracy of the last problem. In addition, the first ten minutes that you check may be more productive than the next ten minutes of checking. So if you wonder if you should forgo the last two problems in order to check your earlier work, you have to redo the experiment anew, measuring both how long it takes you to solve those two problems and the benefits of checking.

This discussion can potentially help you to increase your score. However, there are other strategic considerations to weigh when deciding whether or not to check your work. For example, if the number of mistakes in your tests varies and sometimes you are 100% accurate and you are one problem away from your goal to get to AIME, it is more profitable to go for the last problem and hope for the best. I will discuss the strategic considerations for AMC some other time.